Find the smallest positive \(N \) such that
\(N\equiv3 (mod 4) \)
\(N\equiv2 (mod 5) \)
\(N\equiv6 (mod 7) \)
N mod 4 = 3 N mod 5 = 2 N mod 7 = 6
4A+3=5B+2=7C+6 A = 6, B = 5, C = 3 Therefore: 4 x 6 + 3 = 27
Since LCM of [4, 5, 7] = 140, then: N =140D + 27, where D =0, 1, 2, 3.......etc.
+279 
